Question

A `"25.0-mL"` sample of `"0.150-mol L"^(-1)` hydrofluoric acid is titrated with a `"0.150-mol L"^(-1)` `"NaOH"` solution. What is the `"pH"` after `"26.0 mL"` of base is added?

The `K_a` of hydrofluoric acid is `3.5 xx 10^(-4)`

I know the answer is 11.74 but I'm not sure how to get this (please show steps)

Answer 1

Here's what I got.

Explanation:

You know that hydrofluoric acid, `"HF"`, a weak acid, and sodium hydroxide, which I'll represent as `"OH"^(-)` due to the fact that this compound is a strong base, react in a `1:1` mole ratio to produce fluoride anions.

`color(white)(overbrace(color(black)("HF"_ ((aq))))^(color(blue)("1 mole consumed"))) + color(white)(overbrace(color(black)("OH"_ ((aq))^(-)))^(color(blue)("1 mole consumed"))) -> color(white)(underbrace(color(black)("F"_ ((aq))^(-)))_ (color(blue)("1 mole produced"))) + "H"_ 2"O"_ ((l))`

Use the molarities and the volumes of the two solutions to figure out how many moles of hydrofluoric acid and of hydroxide anions you're mixing.

`25.0 color(red)(cancel(color(black)("mL solution"))) * "0.150 moles HF"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.00375 moles HF"`

`26.0 color(red)(cancel(color(black)("mL solution"))) * "0.150 moles OH"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.00390 moles OH"^(-)`

Now, notice that you have more moles of hydroxide anions than of hydrofluoric acid. Since the two react in a `1:1` mole ratio, you can say that hydrofluoric acid will act as the limiting reagent here, i.e. it will be completely consumed before all the moles of hydroxide anions added to the solution will get the chance to react.

After the reaction is complete, the resulting solution will contain

`"0.00375 moles " - " 0. 00375 moles" = "0 moles HF"`

The hydrofluoric acid will be completely consumed!

`"0.00390 moles " - " 0.00375 moles" = "0.000150 moles OH"^(-)`

The hydroxide anions are in excess, so the resulting solution will contain the hydroxide anions that do not react with the hydrofluoric acid.

The reaction will also produce `0.00375` moles of fluoride anions, which act as a weak base in aqueous solution, but the fact that the solution still contains excess hydroxide anions means that you can ignore the contribution of the fluoride anions to the total concentration of hydroxide anions.

The total volume of the resulting solution will be

`"25.0 mL + 26.0 mL = 51.0 mL"`

This means that the concentration of the hydroxide anions in the resulting solution will be--don't forget to convert the volume of the solution to liters!

`["OH"^(-)] = "0.000150 moles"/(51.0 * 10^(-3) quad "L") = "0.0029412 mol L"^(-1)`

As you know, an aqueous solution at `25^@"C"` has

`"pH + pOH = 14"`

Since

`"pOH" = - log(["OH"^(-)])`

you can say that the `"pH"` of the solution is given by

`"pH" = 14 - [-log(["OH"^(-)])]`

Which is equal to

`"pH" = 14 + log(["OH"^(-)])`

Plug in your value to find

`"pH" = 14 + log(0.0029412) = color(darkgreen)(ul(color(black)(11.47)))`

I'll leave the answer rounded to two decimal places, but keep in mind that you have three sig figs for your values, so you could give the `"pH"` as

`"pH" = 11.469`